Let $n$ be an integer. If the tens digit of $n^2$ is 7, what is the units digit of $n^2$? 
Let $n$ be an integer. If the tens digit of $n^2$ is 7, what is the units digit of $n^2$?

So $n^2 \equiv 7 \pmod{100}$? If this is the case then this can be written as $n^2 = 100k +7$, where $k \in \Bbb Z.$
Here one can see that no matter what the choice of $k$, the units digit will be $7$. Thus $n^2 \equiv 7 \pmod{10}.$ However this was wrong. The correct answer is $\textbf{6}.$
What am I doing wrong here? It seems that $n^2 \equiv 7 \pmod{100}$ doesn't hold. If the tens digit is $7$ should I have that $n^2 \equiv 7k \pmod{100}$, where $k$ represents the unit digit of $70$ and not a multiplication?
 A: You are correct that $n^2\equiv7\bmod100$ does not hold, but rather $n^2\equiv70+k$.
To be a square, the last two digits have to have remainder $0$ or $1$ when divided by $4$
and remainder $0, 1, $ or $4$ when divided by $5$.
Look at the numbers from $70$ to $79$, and figure out which one satisfies those
to figure out what the last digit of $n^2$ must be.
A: Hint You are looking for adigit $k$ such that
$$n^2 \equiv 70+ k \pmod{100}$$
By the Chinese Remainder Theorem this is equivalent to
$$n^2 \equiv 2+ k \pmod{4}\\
n^2 \equiv  k-5\pmod{25}
$$
The quadratic residues modulo $4$ are $0,1$, therefore $k \in \{ 2,3, 6,7 \}$. You have now to figure for which of those $k-5$ is a quadratic residue modulo $25$.
A: Any number that is a square mod $100$ is necessarily a square both mod $4$ and mod $5$, which is to say $0$ or $1$ mod $4$ and $0$, $1$, or $4$ mod $5$. The only number in the $70$s that satisfies both criteria is $76$.
A: Hint
Look at the order of these solutions:
$\quad 24^2 = 576$
$\quad 74^2 = 5476$
$\quad 26^2 = 676$
$\quad 76^2 = 5776$
A: The tens digit is $7$, not the units and you want to find the unit digit.
So if the unit digit is $x$ then the number ends with $7x$ and $\pmod{100}$ what you are trying to say is $n^2 \equiv 70 + x\pmod {100}$.
The way I would do this is let $n= 10k + a$ where $a,k$ are single digits.  And the hundreds place don't affect the last two digits we might as well assume $n$ has only two digits.
$n^2 = 100k^2 + 20ak + a^2 = 100m + 70 + x$.
As $7$ is odd but $2ak$ is even so $a^2$ must be two digits and we carried an odd digit.  if $a = 0,1,2,3,4,5,6,7,8,9$ then $a^2 = 0,1,4,9,14,25,36,49,64,81$..  So $a = 4$ or $6$.
So if $a = 4$ and $n=10k +4$ then $n^2 = 100k^2 + 80k+ 16$ and if $a=6$ and $n=10k+6$ then $n62 = 100k^2 + 120k + 36$.  In either event $x = 6$.
We can have $8k +1\equiv 7$ and $k = 2,7$ or we can have $2k+3 \equiv 7$ and $k=2,7$.  Notice $24,26,74,76$ when squared all end with $76$.
